Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x-3y &= 1 \\ 7x-6y &= 2\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $7x = 6y+2$ Divide both sides by $7$ to isolate $x$ $x = {\dfrac{6}{7}y + \dfrac{2}{7}}$ Substitute this expression for $x$ in the first equation. $6({\dfrac{6}{7}y + \dfrac{2}{7}}) - 3y = 1$ $\dfrac{36}{7}y + \dfrac{12}{7} - 3y = 1$ Simplify by combining terms, then solve for $y$ $\dfrac{15}{7}y + \dfrac{12}{7} = 1$ $\dfrac{15}{7}y = -\dfrac{5}{7}$ $y = -\dfrac{1}{3}$ Substitute $-\dfrac{1}{3}$ for $y$ in the top equation. $6x-3( -\dfrac{1}{3}) = 1$ $6x+1 = 1$ $6x = 0$ $x = 0$ The solution is $\enspace x = 0, \enspace y = -\dfrac{1}{3}$.